Problem 13
Question
In Exercises, factor the polynomial. If the polynomial is prime, state it. $$ x^{2}-x y-6 y^{2} $$
Step-by-Step Solution
Verified Answer
The given polynomial is prime. Factored form remains unchanged: \(x^2 - xy - 6y^2\).
1Step 1: Identify the type of polynomial
First, we notice that the polynomial given is of the form ax^2 + bxy + cy^2, which is a quadratic trinomial.
2Step 2: Finding the constants
We now need to find two constants, p and q that fulfil the following criteria:
1. p * q = -6y^2
2. p + q = -xy
3Step 3: Multiplicative Constants
To ensure that p * q = -6y^2, we write the product of the possible pairs of numbers:
1. (1 * -6y^2)
2. (-1 * 6y^2)
3. (2 * -3y^2)
4. (-2 * 3y^2)
Now let's find the pair which adds up to -xy.
4Step 4: Additive constants
For each of the pairs above, check if they add up to -xy.
1. (1) + (-6y^2) = -6y^2 + 1 ≠ -xy
2. (-1) + (6y^2) = 6y^2 - 1 ≠ -xy
3. (2) + (-3y^2) = -3y^2 + 2 ≠ -xy
4. (-2) + (3y^2) = 3y^2 - 2 ≠ -xy
We notice that none of the pairs add up to -xy.
5Step 5: State if the polynomial is prime
Since we're unable to find any pairs (p, q) that fulfil the required criteria, the given polynomial is prime. Therefore, the factored form of the polynomial is unchanged:
\(
x^{2}-x y-6 y^{2}
\)
Key Concepts
Quadratic TrinomialPrime PolynomialMathematical Constants
Quadratic Trinomial
A quadratic trinomial is a polynomial expression that consists of three terms. These terms generally include a variable raised to the power of two, such as \( ax^2 + bxy + cy^2 \). This form makes it recognizable as a quadratic because the highest exponent of the variable is two. Quadratic trinomials are very common in algebra and often require factoring to simplify or solve.
Factoring a quadratic trinomial can involve various strategies, depending on its structure:
Factoring a quadratic trinomial can involve various strategies, depending on its structure:
- Perfect square trinomials can be factored into the square of a binomial.
- Other trinomials might need techniques such as factoring by grouping or applying the quadratic formula to solve.
Prime Polynomial
A prime polynomial is similar to a prime number. Just as prime numbers have no divisors other than 1 and themselves, prime polynomials cannot be factored into simpler, non-trivial polynomials over the integers. This means that no matter how you try, you won't be able to find factors that multiply to the polynomial other than 1 and the polynomial itself.
Identifying a prime polynomial requires testing potential factor combinations and confirming none work. In our example, none of the tried pairs satisfied both the multiplicative and the additive conditions needed to factor the quadratic trinomial \( x^2 - xy - 6y^2 \). Therefore, it remains unfactored and is classified as prime. Recognizing when a polynomial is prime is essential, as it can save time and effort in algebraic problem-solving.
Identifying a prime polynomial requires testing potential factor combinations and confirming none work. In our example, none of the tried pairs satisfied both the multiplicative and the additive conditions needed to factor the quadratic trinomial \( x^2 - xy - 6y^2 \). Therefore, it remains unfactored and is classified as prime. Recognizing when a polynomial is prime is essential, as it can save time and effort in algebraic problem-solving.
Mathematical Constants
In the context of polynomial factoring, constants refer to fixed numbers that help in simplifying or factoring expressions. They remain unchanged across operations. When dealing with polynomials like the one in our exercise, constants play a crucial role in determining how the expression can be split or factored.
For instance, to factor the polynomial \( x^2 - xy - 6y^2 \), we looked for constants \( p \) and \( q \) that could balance both multiplication and addition based on specific criteria:
For instance, to factor the polynomial \( x^2 - xy - 6y^2 \), we looked for constants \( p \) and \( q \) that could balance both multiplication and addition based on specific criteria:
- Their product needed to match the last term, \( -6y^2 \).
- Their sum needed to match the middle term, \( -xy \).
Other exercises in this chapter
Problem 13
Solve the given equation. $$ \frac{3}{5}(k+1)=\frac{1}{4}(2 k+3) $$
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Rewrite the number without using exponents. $$ \left(a b^{2}\right)^{0}, \text { where } a, b \neq 0 $$
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Indicate whether the statement is true or false. Every natural number is an integer.
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Perform the indicated operations and simplify. $$ (2 x+3)+(4 x-6) $$
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